Compute $\lim_{x\to\infty}x\;\left[\left(1+\frac{1}{x}\right)^x-e\right]$ [duplicate]

How can I compute the following limit, and is there a general method to resolve problems of this type?

$$\lim_{x\to\infty}x\left[\left(1+\frac{1}{x}\right)^x-e\right]$$

marked as duplicate by YuiTo Cheng, Michael Rozenberg calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 4 at 5:13

• $(+\infty)^2 =+\infty$ – Paolo Leonetti Oct 10 '16 at 14:48
• I think inside need to be $x[(1+1/x)^x-e]$ – student forever Oct 10 '16 at 14:50
• I agree, but we cannot also try to interpret what he wants to ask – Paolo Leonetti Oct 10 '16 at 14:52
• @studentforever You may be right as $$(x+\frac{1}{x})^x=e^{x(log(x(1+\frac{1}{x^é})))}=e^{x(log(x)+log((1+\frac{1}{x^2}))))}$$ tends to infinity. Wait for the OP reaction. – Duchamp Gérard H. E. Oct 10 '16 at 14:53
• Yeah fixed the typo... – Joshua Benabou Oct 10 '16 at 15:04

Using the Taylor Series for $\log(1+x)$, we get \begin{align} x\log\left(1+\frac1x\right) &=x\left(\frac1x-\frac1{2x^2}+O\left(\frac1{x^3}\right)\right)\\ &=1-\frac1{2x}+O\left(\frac1{x^2}\right) \end{align} Therefore, using the Taylor Series for $e^x$, we get \begin{align} x\left[\left(1+\frac1x\right)^x-e\right] &=x\left[e^{1-\frac1{2x}+O\left(\frac1{x^2}\right)}-e\right]\\ &=xe\left[e^{-\frac1{2x}+O\left(\frac1{x^2}\right)}-1\right]\\ &=xe\left[-\frac1{2x}+O\left(\frac1{x^2}\right)\right]\\ &=-\frac e2+O\left(\frac1x\right) \end{align} Thus, $$\lim_{x\to\infty}x\left[\left(1+\frac1x\right)^x-e\right]=-\frac e2$$
hint:$$\lim_{x\to\infty}{x\left(\left(1+\frac{1}{x}\right)^x-e\right)}=\lim_{x\to\infty}\frac{{\left(1+\frac{1}{x}\right)^x-e}}{\frac{1}{x}}=$$ then apply L'Hospital rule